A water pipe tapers down from an initial radius of R 1 = 0.21 m to a final radiu

Question

A water pipe tapers down from an initial radius of R1 = 0.21 m to a final radius of R2 = 0.11 m. The water flows at a velocity v1 = 0.83 m/s in the larger section of pipe.

1)

What is the volume flow rate of the water?

m3/s

2)

What is the velocity of the water in the smaller section?

m/s

3)

Using this water supply, how long would it take to fill up a swimming pool with a volume of V = 165 m3? (give your answer in minutes)

min

4)

The water pressure in the center of the larger section of the pipe is P1 = 253250 Pa. Assume the density of water is 103 kg/m3.

What is the pressure in the center of the smaller section of the pipe?

Pa

5)

If the pipe was turned vertical and the volume flow rate in the larger section is kept the same, which answers would change?

the speed of water in the smaller section

the volume flow rate in the smaller section

the pressure in the smaller section

PLEASE, CLEARLY WORK IT OUT CORRECTLY..THNKS

Explanation / Answer

1) Volume flow rate Q= VA = 0.83* pi*0.21^2 = 0.115 m^3/s

2)V2 = Q/A2 = 0.115/(pi*0.11*0.11) = 3.025 m/s

3) time = 165/0.115 = 1435 s = 23.9 minutes

4)By Bernoullis equation

P2 = P1 + 0.5 rho (v1^2 -v2^2)

= 253250 + 0.5*1000* (0.83^2 -3.025^2)

= 249019 Pa

5) Only pressure will change

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